Solution
= Solution
After these augmentations, the vertices reachable from $S$ in the <residual network> are
$$
A=\{S,a,c\}.
$$
The edges leaving this set are $S\to b$, $a\to d$, and $c\to e$, with capacities $3$, $1$, and $4$. Hence
$$
c(A,V\setminus A)=3+1+4=8.
$$
The flow in part (i) has the same value, so the <max-flow min-cut theorem> proves that it is maximum and that this cut is minimum.